Minimum excitation potential of Bohr's first orbit in hydrogen atom is

3.6 V

10.2 V

13.6 V

3.4 V

Solution

10.2 V

The energy of an electron in Bohr's orbit of hydrogen atom is given by the expression

$E_{n} = - \frac{2 π^{2} {me}^{2} Z^{2}}{n^{2} h^{2} {(4 {πε}_{o})}^{2}}$

$E_{n} = - 13.6 \frac{Z^{2}}{n^{2}} eV$

Since Z = 1 for hydrogen above equation can be further simplified to

E_n = $\frac{- 13.6}{n^{2}} eV$

From the relation

$∆ E = 13.6 eV - \frac{13.6 eV}{n^{2}}$

= $13.6 - \frac{13.6 eV}{{(2)}^{2}}$

ΔE = 10.2 eV

Therefore, excitation potential

= $\frac{10.2}{e} eV$

= 10.2 V

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